Techniques for Solving Solvable Equations via Galois Theory

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[9-2] Pitfalls in solving equations

The reason we were unable to compute \( A_2^{-1}\) in the previous section appears to be the following.

“The minimal polynomial \(V(x)=g_0(x)\) is not irreducible over the base field \(F_0\).”

Indeed, the \(V(x)\) in (1.5) is irreducible over the rational field \(Q\).
However, algebra textbooks define an “algebraic solution” as follows.

“Starting from the coefficients of \(f(x)\), one obtains the roots by the four arithmetic operations and radicals (assuming the \(n\)th roots of unity are known).”

In other words, when solving equations by Galois theory, it is assumed from the outset that the base field \(F_0\) already contains the relevant roots of unity—in this case the cube roots of unity. Hence the base field should be \(F_0=Q(\omega)\), not \(F_0=Q\).

We must therefore check whether the polynomial \(V(x)=x^6+108\), which we had taken as the minimal polynomial of \(v\) in (1.5), is actually irreducible over the base field \(F_0=Q(\omega)\). Using the Trager Algorithm to test irreducibility over \(F_0=Q(\omega)\), we find that \(V(x)\) factors as in (2.1). We thus define the minimal polynomial \(g_0(x)\) of \(v\) as in (2.2).

\begin{align} V(x)&=x^6+108=(x^3-12\omega-6)(x^3+12\omega+6) \\ \notag \\ \therefore \ g_0(x) &\equiv x^3-12\omega-6 \\ \end{align}

Next, we factor \(f(x)\) over the number field \(F_0(v)=Q(\omega,v)\) generated by this minimal polynomial \([ \ g_0(v)=0 \ ]\). Again we use the Trager Algorithm. The three roots of \(f(x)\) are then obtained as follows.

\begin{align} &\alpha =-\frac{v( \omega +2)}{3}, \quad \beta =\frac{v(2 \omega +1)}{3}, \quad \gamma =-\frac{v( \omega -1)}{3} \\ \notag \\ &v=\alpha+2\beta+3\gamma \\ &\qquad \Downarrow \notag \\ \end{align} \begin{align} &\left\{ \begin{array}{l} \sigma_{1}(v)=v_{1}=v & &\sigma_{2}(v)=v_{2}=v(\omega+1) & &\sigma_{3}(v)=v_{3}=-v\omega \\ \sigma_{4}(v)=v_{4}=-v(\omega+1) & &\sigma_{5}(v)=v_{5}=v\omega & &\sigma_{6}(v)=v_{6}=-v \\ \end{array} \right.\\ \end{align}

\begin{align} &\qquad \Downarrow \notag \\ & g_0(v_1)=g_0(v_4)=g_0(v_5)=0 \qquad g_0(v_2)=g_0(v_3)=g_0(v_6)=-24\omega-12 \\ &\qquad \Downarrow \notag \\ \end{align} \begin{align} &\rho_{1}(v)=v_1=v & &\rho_{4}(v)=v_4=-v(\omega+1) & &\rho_{5}(v)=v_5=v\omega & \end{align}

Computation of the cyclic cubic extension \(F_1/F_0\). We record only the results.

Step1 LRT (Lagrange Resolvent transformation)

\begin{align} &Gal(F_1/F_0)=A_3/e \equiv C_3=\{\rho_1,\rho_4,\rho_5\} \notag \\ \notag \\ &\left\{ \begin{array}{l} h_0=\rho_1(x-v)=(x-v_1)=x-v \\ h_1=\rho_4(x-v)=(x-v_4)=x+v(\omega+1) \\ h_2=\rho_5(x-v)=(x-v_5)=x-v\omega \\ \end{array} \right. \\ \end{align} \begin{align} \notag \\ &\begin{bmatrix} t_0 \\ t_1 \\ t_2 \end{bmatrix} =\frac{1}{3} \begin{bmatrix} 1&1&1 \\ 1&\omega&\omega^2\\ 1&(\omega^2)&(\omega^2)^2\\ \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \\ h_2 \end{bmatrix} = \begin{bmatrix} x\\ -v\\ 0\\ \end{bmatrix} \\ \end{align}

Step2 Binomial \(B_2(x)\) and the new adjoined element \(a_2\)

\begin{align} &\left\{ \begin{array}{l} t_0 \ \in \ F_0[x] \\ t_1 \ \in \ F_0(v) \\ t_2 \ \in \ F_0 \\ \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} t_1^3=A_1=-12\omega-6 \in \ F_0 \\ B_1(x)=x^3-A_1=0 \\ a_1=\sqrt[3]{A_1} \ \in \ F_0(a_1) \equiv \ F_1 \\ \end{array} \right. \\ \end{align}

Step3 ILRT (Inverse Lagrange Resolvent transformation)

\begin{align} &\begin{bmatrix} \tilde{h_0} \\ \tilde{h_1} \\ \tilde{h_2} \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 1&\omega^2&(\omega^2)^2\\ 1&\omega&(\omega^2)\\ \end{bmatrix} \cdot \begin{bmatrix} t_0 \\ \tilde{t_1} \\ \tilde{t_2} \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 1&\omega^2&(\omega^2)^2\\ 1&\omega&(\omega^2)\\ \end{bmatrix} \cdot \begin{bmatrix} x \\ a_1 \\ 0 \end{bmatrix} = \begin{bmatrix} x+a_1 \\ x+a_1\omega^2 \\ x+a_1\omega \\ \end{bmatrix} \\ \notag \\ &\therefore \quad g_1(x) \equiv \tilde{h_0} =x+a_1\ \in \ F_1[x] \\ \end{align}

\begin{align} & g_1(x)=0 \quad \rightarrow \quad v=-a_1 \notag \\ &\qquad \Downarrow \notag \\ &\therefore \ \alpha =\frac{a_1( \omega +2)}{3}, \quad \beta =-\frac{a_1(2 \omega +1)}{3}, \quad \gamma =\frac{a_1( \omega -1)}{3} \\ \end{align}

The previous examples happened to work smoothly because in each case the minimal polynomial was irreducible over the chosen base field \(Q(\omega)\) or \(Q(\zeta)\). Going forward, when determining a minimal polynomial it is essential to check irreducibility over the base field.

In the cyclic extension above we invoked the Trager Algorithm twice—once to factor the minimal polynomial and once to factor the original equation—but omitted details. In the next section, using a slightly more complicated equation, we will show a worked example that both reviews the Trager Algorithm and illustrates the actual computations involved.

[Supplement] Below are two tables. They list, for solvable equations, the degrees of the minimal polynomials in a simple extension \(Q(v)\) and in the extension obtained by adjoining a primitive root of unity in advance, \(Q(v,\zeta)\) or \(Q(v,\omega)\). Take care with cases where \(deg(g(v)) \neq deg(g(v,\zeta))\) and with the equations highlighted in yellow. An even more puzzling phenomenon occurs for the equation \(f_7(x)\). Its minimal polynomial \(g(v)\) is not irreducible over \(Q(v,\zeta)\), yet proceeding “as is” still yields correct computations. Of course, using the minimal polynomial \(g(v,\zeta)\) over \(Q(v,\zeta)\) also leads to the correct answer. Why does this happen? Consider it carefully.

Quintic equations and degrees of their minimal polynomials \(Z=\zeta^{4}+\zeta^{3}+\zeta^{2}+\zeta +1\)
\(f_i(x)\) \(equation\) \(deg(g(v))\) \(deg(g(v,\zeta))\) \(Gal(Q(v,\zeta)/Q(\zeta)) \)

\({f_1(x)}\) \(x^5+x^4+2x^3+4x^2+x+1\) \(20\) \(20\) \(F_{20}\)

\({f_2(x)}\) \(x^5-5x^3+5x+6\) \(20\) \(10\) \(D_{5}\)

\({f_3(x)}\) \(x^5-5x^3+5x-3\) \(20\) \(5\) \(C_{5}\)

\({f_4(x)}\) \(x^5-3\) \(20\) \(5\) \(C_{5}\)

\({f_5(x)}\) \(x^5-x^3-2x^2-2x-1\) \(10\) \(10\) \(D_{5}\)

\({f_6(x)}\) \(x^5-10x^3+5x^2+10x+1\) \(5\) \(5\) \(C_{5}\)

\({f_7(x)}\) \(x^5+15x+12\) \(20\) \(10\) \(D_{5}\)

Quintic equations and degrees of their minimal polynomials \(Z=\zeta^{4}+\zeta^{3}+\zeta^{2}+\zeta +1\)
\(f_i(x)\)	\(equation\)	\(deg(g(v))\)	\(deg(g(v,\zeta))\)	\(Gal(Q(v,\zeta)/Q(\zeta)) \)
\({f_1(x)}\)	\(x^5+x^4+2x^3+4x^2+x+1\)	\(20\)	\(20\)	\(F_{20}\)
\({f_2(x)}\)	\(x^5-5x^3+5x+6\)	\(20\)	\(10\)	\(D_{5}\)
\({f_3(x)}\)	\(x^5-5x^3+5x-3\)	\(20\)	\(5\)	\(C_{5}\)
\({f_4(x)}\)	\(x^5-3\)	\(20\)	\(5\)	\(C_{5}\)
\({f_5(x)}\)	\(x^5-x^3-2x^2-2x-1\)	\(10\)	\(10\)	\(D_{5}\)
\({f_6(x)}\)	\(x^5-10x^3+5x^2+10x+1\)	\(5\)	\(5\)	\(C_{5}\)
\({f_7(x)}\)	\(x^5+15x+12\)	\(20\)	\(10\)	\(D_{5}\)

Cubic equations and degrees of their minimal polynomials \(\Omega=\omega^{2}+\omega +1\)
\(f_i(x)\) \(equation\) \(deg(g(v))\) \(deg(g(v,\omega))\) \(Gal(Q(v,\omega)/Q(\omega)) \)

\({f_8(x)}\) \(x^3+3x+1\) \(6\) \(6\) \(S_{3}\)

\({f_{9}(x)}\) \(x^3-2\) \(6\) \(3\) \(A_{3} \ (\cong C_3)\)

\({f_{10}(x)}\) \(x^3-3x+1\) \(3\) \(3\) \(A_{3} \ ( \cong C_3)\)

Cubic equations and degrees of their minimal polynomials \(\Omega=\omega^{2}+\omega +1\)
\(f_i(x)\)	\(equation\)	\(deg(g(v))\)	\(deg(g(v,\omega))\)	\(Gal(Q(v,\omega)/Q(\omega)) \)
\({f_8(x)}\)	\(x^3+3x+1\)	\(6\)	\(6\)	\(S_{3}\)
\({f_{9}(x)}\)	\(x^3-2\)	\(6\)	\(3\)	\(A_{3} \ (\cong C_3)\)
\({f_{10}(x)}\)	\(x^3-3x+1\)	\(3\)	\(3\)	\(A_{3} \ ( \cong C_3)\)

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